# 2007 iTest Problems/Problem 21

## Problem

James writes down fifteen 1's in a row and randomly writes + or - between each pair of consecutive 1's. One such example is $$1+1+1-1-1+1-1+1-1+1-1-1-1+1+1.$$ What is the probability that the value of the expression James wrote down is $7$? $\text{(A) }0\qquad \text{(B) }\frac{6435 }{2^{14}}\qquad \text{(C) }\frac{6435 }{2^{13}}\qquad \text{(D) }\frac{429}{2^{12}}\qquad \text{(E) }\frac{429}{2^{11}}\qquad \text{(F) }\frac{429}{2^{10}}\qquad \text{(G) }\frac{1}{15}\qquad \text{(H) }\frac{1}{31}\qquad$ $\text{(I) }\frac{1}{30}\qquad \text{(J) }\frac{1}{29}\qquad \text{(K) }\frac{1001 }{2^{15}}\qquad \text{(L) }\frac{1001 }{2^{14}}\qquad \text{(M) }\frac{1001 }{2^{13}}\qquad \text{(N) }\frac{1}{2^{7}}\qquad \text{(O) }\frac{1}{2^{14}}\qquad \text{(P) }\frac{1}{2^{15}}\qquad$ $\text{(Q) }\frac{2007}{2^{14}}\qquad \text{(R) }\frac{2007}{2^{15}}\qquad \text{(S) }\frac{2007}{2^{2007}}\qquad \text{(T) }\frac{1}{2007}\qquad \text{(U) }\frac{-2007}{2^{14}}\qquad$

## Solution

In James’s expression, he is adding or subtracting $14$ ones to one. Since the wanted result is $7$, he needs to add a total of $6$ to $1$. This is achieved when he writes $10$ plus signs and $4$ minus signs, and there are $\binom{14}{10} = 1001$ possible ways to do that. Since the total number of ways to put plus signs and minus signs between consecutive ones is $2^{14}$, the probability of getting a $7$ is $\boxed{\textbf{(L) } \frac{1001}{2^{14}}}$.

## See Also

 2007 iTest (Problems) Preceded by:Problem 20 Followed by:Problem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4
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